Does the generic wave equation imply the universe has a Minkowski spacetime?

Question

https://en.wikipedia.org/wiki/Wave_equation#Introduction
Does the wave equation
$$
(partial_t{}^2 -c^2 nabla^2) u=0
$$
imply the metric of the universe is Minkowski (-like)
$$
g= (+,-c^2, -c^2 ,-c^2 )
$$
See the d'Alembert operator https://en.wikipedia.org/wiki/D%27Alembert_operator
I came across the wave equation in my classical mechanics course in an effort to reproduce coupled oscillator behavior from an action and was flabbergasted when I just started using the d'Alembert operator without even realizing.

user285793 · Answer

This wave equation merely asserts something about the scalar field $u$. A scalar field could violate this wave equation but obey the Klein-Gordon equation instead, and that would not change the structure of spacetime.
It's also possible for the universe not to be Minkowski but for this wave equation nevertheless to be the one obeyed by some field. In fact, the universe isn't Minkowskian -- it's only locally, approximately Minkowskian.
What we can say is that this wave equation is connected to the structure of spacetime in that, if spacetime is locally Minkowski, it's the only Lorentz-invariant second-order wave equation you can write for a scalar field if the only tool at your disposal is the derivatives (and not, say, a mass).

Does the generic wave equation imply the universe has a Minkowski spacetime?

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